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Question
Find a quadratic equation whose roots are sin 15° and cos 15°
Solution
sin 15° = sin(45° – 30°)
= sin 45°. cos 30° – cos 45°. sin 30°
= `1/sqrt(2) * sqrt(3)/2 - 1/sqrt(2) * 1/2`
sin 15° = `1/(2sqrt(2)) (sqrt(3) - 1)` ......(1)
cos 15° = cos(45° – 30°)
= cos 45° . cos 30° + sin 45° . sin 30°
= `1/sqrt(2) * sqrt(3)/2 + 1/sqrt(2) * 1/2`
sin 15° = `1/(2sqrt(2)) (sqrt(3) + 1)` ......(2)
The quadratic whose roots cos 15° and sin 15° is
x2 – (cos 15° + sin 15°)x + (cos 15°) (sin 15°) = 0 ......(3)
cos 15° + sin 15° = `1/(2sqrt(2)) (sqrt(3) + 1) + 1/(2sqrt(2)) (sqrt(3) - 1)`
= `1/(2sqrt(2)) (sqrt(3) + 1 + sqrt(3) - 1)`
= `(2sqrt(3))/(2sqrt(2))`
= `sqrt(3)/sqrt(2)`
= `sqrt(3)/sqrt(2) xx sqrt(2)/sqrt(2)`
= `sqrt(6)/2`
(cos 15°) (sin 15°) = `1/(2sqrt(2)) (sqrt(3) + 1) * 1/(2sqrt(2)) (sqrt(3) - 1)`
= `1/(4(2)) * (3 - 1)`
= `2/8`
= `1/4`
Substituting in equation (3) we have
`x^2 - sqrt(6)/2x + 1/4` = 0
`4x^2 - 2sqrt(6)x + 1` = 0
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